Wikipedia:Reference desk/Archives/Mathematics/2023 June 18

= June 18 =

What anything, more specific can as simple/general as possible, should we know about a given function $$f$$, if for concluding that $$\frac{f(x)}{x}$$ is known to be a one-to-one correspondence?
I have no objection to stipulating that $$f$$ is from the set of positive numbers to iteself. Want to assume also that $$f$$ is monotonic? No ojection. Continuous? No objection. Differntiable? No objection. 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 04:42, 18 June 2023 (UTC)


 * If $$g: x\mapsto f(x)/x$$ is one-to-one and continuous on the positive reals, it is strictly monotonic. And a continuous strictly monotonic function is 1-1. So any function $$f$$ that can be written as $$f(x)=g(x)x$$ for a strictly monotonic $$g$$ will work. —Kusma (talk) 08:49, 18 June 2023 (UTC)
 * And under the assumption that $$f$$ is continuous, this is a most general form for $$f$$. --Lambiam 09:11, 18 June 2023 (UTC)
 * $$f(x) = g(x)x$$ where $$g$$ is (continuous and) strictly monotonic can be defined precisely as $$f$$ which is (continuous and) strictly monotonic in one direction over $$\mathbb{R}^{-}$$, (continuous and) strictly monotonic in the other direction over $$\mathbb{R}^{+}$$, and $$0$$ at $$0$$. GalacticShoe (talk) 13:03, 18 June 2023 (UTC)
 * In general, an injection $$g : \mathbb{R} \rightarrow \mathbb{R}$$ is a bijection $$g : \mathbb{R} \rightarrow g(\mathbb{R})$$, where $$g(\mathbb{R}) \subseteq \mathbb{R}$$ is uncountable. Moreover, if $$D \subseteq \mathbb{R}$$ and $$g : \mathbb{R} \rightarrow D$$ is bijective, then naturally $$g : \mathbb{R} \rightarrow \mathbb{R}$$ is injective. So the most general form is $$f(x) = xg(x)$$ where $$g$$ is bijective from $$\mathbb{R}$$ to some uncountable subset of $$\mathbb{R}$$. Of course, this is a very general (and redundant) form and doesn't seem to have any nice properties, but that's at least partially because there's not a lot of structure to be gleaned from the one injectivity requirement. There are some very pathological examples one can construct here, mostly from bizarre $$g : \mathbb{R} \rightarrow D$$. Consider, for example, a bijection between the real numbers and the Cantor set. GalacticShoe (talk) 13:23, 18 June 2023 (UTC)
 * As an example of a bijection between the real numbers and the Cantor set, one can construct a bijection $$f : \mathbb{R} \rightarrow (0, 1)$$ defined by $$f(x) = 1/(1+\exp(-x))$$, and then one can construct a bijection $$g : (0, 1) \rightarrow \mathcal{C}$$ defined by taking $$x \in (0, 1)$$, writing out its binary expansion, converting all $$1$$s to $$2$$s, then converting it from trinary to decimal, finally yielding the function $$h = g \circ f$$ as a bijection $$h : \mathbb{R} \rightarrow \mathcal{C}$$, which looks like this. The function that results from $$xh(x)$$ looks like this, and does seem rather pathological for a function which, divided by $$x$$, is injective. For one, it looks like there should be quite a few lines that intersect the function in more than one place, but that's just a consequence of both the graphics used (Matlab draws the curve as a continuous line), as well as the gaps in the Cantor set. GalacticShoe (talk) 14:15, 18 June 2023 (UTC)

OP's clarification: Sorry for my mistake, which I've just fixed on the header: I meant just the opposite: For a given function $$f,$$ I would like to conclude that $$g: x\mapsto f(x)/x$$ is one-to-one, by a previous simple information about $$f$$ (that can be srtictly monotonic or continuous/differentiable or defined from the set of positive numbers to itself, as you wish), without g being mentioned in that previous information about $$f.$$ 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 12:16, 18 June 2023 (UTC)


 * Note that the previous conversations apply pretty much entirely the other way too, so $$f(x)$$ being equal to $$xg(x)$$ for $$g$$ continuous and strictly monotonic implies $$f$$ being continuous and injective, and $$f(x)$$ being equal to $$xg(x)$$ for bijective $$g : \mathbb{R} \rightarrow D$$ for $$D \subseteq \mathbb{R}$$ implies $$f$$ being injective. GalacticShoe (talk) 13:32, 18 June 2023 (UTC)
 * A sufficient condition is to assume that $$f$$ is differentiable and $$xf'(x)-f(x)$$ is everywhere positive or everywhere negative. For example, any $$f$$ such that $$f(x)>0$$ and $$f'(x)\le 0$$ for every $$x$$. —Kusma (talk) 13:36, 18 June 2023 (UTC)
 * Quick note that the condition on $$f'(x) \leq 0$$ for every $$x$$ should be $$f'(x) \leq 0$$ for $$x \geq 0$$ and $$f'(x) \geq 0$$ for $$x < 0$$; otherwise, you can get, for example, $$f(x) = \exp(-x)$$, $$f'(x) = -\exp(-x)$$, and $$xf'(x) - f(x) = -x\exp(-x) - \exp(-x)$$ which is not everywhere positive/negative. GalacticShoe (talk) 14:28, 18 June 2023 (UTC)
 * I am working only in positive numbers. As we are talking about a function written as $$f(x)/x$$, we can't use $$x=0$$ anyway. —Kusma (talk) 15:59, 18 June 2023 (UTC)
 * Fair point given that we are working with $$f(x)/x$$ specifically, I just thought it worth pointing out that the distinction is necessary in general. GalacticShoe (talk) 17:28, 18 June 2023 (UTC)

Starting anew
Sorry for my mistakes in the previous thread. I thought, I had to strike out some words on the header, and to add other words, but now I realize my previous thought was a second mistake, so I decide to start anew, being as exact as possible, this time:

What I know is the following: Both $$g$$ and $$f: x\mapsto xg(x)$$ are continuous (and even differentiable) injections, defined for every positive number.

[Later addition: Hence their injectivity can be replaced by their strict monotonicity]. $$f$$ is also known to be strictly monotonic.

Besides this information, can I infer something else (anything non obvious) about $$f$$ (rather than about $$g)?$$ 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 16:30, 18 June 2023 (UTC)


 * Any continuous injective function is strictly monotonic. (For a proof, see Proposition 5.7.2 here; I only saw the if direction on Wikipedia.) So the strict monotonicity of $$f(x)$$ already follows from its given continuity and $$g$$ is also known to be strictly monotonic. --Lambiam 07:32, 19 June 2023 (UTC)
 * Thanks. I've just added your information [in brackets] to my original question on this thread. See the third paragraph above. 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk)
 * If both $$g(x)$$ is continuous, then $$xg(x)$$ is automatically continuous as well. In terms of $$g(x)$$ being monotonic, however, it is a necessary (but not sufficient) condition that $$g(x)$$ have no zeroes. If $$g(0) = 0$$, then $$xg(x)$$ is continuous and either always positive or always negative except for at $$0$$, which immediately implies that all neighborhoods of $$0$$ breaks the injectivity constraint. If $$g(x_{0}) = 0$$ for $$x_{0} \neq 0$$ meanwhile, then $$xg(x)$$ is $$0$$ at $$0$$ and $$x_{0}$$, meaning that all combined neighborhoods of $$0$$ and $$x_{0}$$ break the injectivity constraint. GalacticShoe (talk) 15:58, 19 June 2023 (UTC)
 * Beyond $$g$$ being either all-positive or all-negative, there's actually not much one can glean when $$g$$ is just continuous. For example, although the monotone convergence theorem guarantees that $$g$$ has a limit in at least one direction, adding conditions on the limit doesn't seem to make $$xg(x)$$ any more or less likely to be monotonic. For example, since $$x(e^{x} + N)$$ is not monotonic for all $$0 \leq N < 1/e^{2} \approx 0.1353$$, the function $$g(x) = 10Ke^{x} + K$$ has arbitrary limit $$K \neq 0$$, yet yields $$xg(x)$$ that is not monotonic. I have yet to consider if differentiability allows for some better properties, but I'm not particularly hopeful on that front, given that monotonic functions in general already are differentiable everywhere except a set of measure zero. GalacticShoe (talk) 16:30, 19 June 2023 (UTC)
 * OP's response. I think I have gleaned some details about $$f,$$ yet they are too specific, whereas I'm looking for more general (non-obvious) ones.
 * As mentioned above, both $$g$$ and $$f: x\mapsto xg(x)$$ are continuous (and even differentiable) injections. So here are some specific details I can glean about $$f:$$
 * 1. $$f,$$ defined over the set of real numbres, cannot be the specific function $$f(x)=Bx,$$ for any real $$B \ne 0,$$ because even though $$f$$ is a continuous injection as required, $$g$$ is not.
 * 2. $$f,$$ defined over the set of real numbres, cannot be the specific function $$f(x)=x^B,$$ for any odd positive exponential $$B,$$ because even though $$f$$ is a continuous injection as required, $$g$$ is not.
 * 3. $$f,$$ defined over the set of positive numbres, cannot be the specific function $$f(x)=\log_{B}(x),$$ for any positive base $$B \ne 1,$$ because even though $$f$$ is a continuous injection as required, $$g$$ is not.
 * 4. $$f,$$ defined over the interval $$(0,1),$$ cannot be the specific function $$f(x)=\arcsin\sqrt{x},$$ because even though $$f$$ is a continuous injection as required, $$g$$ is not.
 * So, could you fill in the blanks?
 * 5. $$f$$ cannot be any function having the following general (non-obvious) property: ____________________, because even though $$f$$ is a continuous injection as required, $$g$$ is not.
 * Note that by "non-obvious" property, I intend to exclude any property which $$f$$ is obviously not permitted to have, like: "$$f(x)=xg(x)$$ for some continuous (and even differentiable) function $$g$$ that is not injective while $$f$$ is".
 * 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 19:37, 19 June 2023 (UTC)
 * If $$f$$ [defined over the set of positive numbres (OP's addition) ] is differentiable, the graph of its derivative $$f'$$ cannot cross the positive x-axis, that is, the equation $$f'(x)=0$$ has no positive solutions in $$x$$. But the equation $$f'(x)=g(x)$$ can also not have any positive solutions. Some examples. For $$f(x)=Bx$$ we have $$f'(x)=g(x)$$ everywhere. For $$f(x)=\log_B(x),B>0,B\ne 1,$$ we have $$f'(e)=g(e)=e^{{-}1}\log_Be.$$ For $$f(x)=B^x,B>1,$$ we have $$f'(\log_Be)=g(\log_Be)=e\log B.$$ So none of these are OK. But for $$f(x)=x^B,B\ne0,B\ne 1,$$ we have $$f'(x)=Bx^{B-1}\ne g(x)=x^{B-1},$$ and for $$f(x)=x\exp x,$$ we have $$f'(x)=(x+1)\exp x\ne g(x)=\exp x,$$ so these are fine. Also, for $$f(x)=B^x,00,$$ so $$f(x)=B^x$$ is OK provided that $$0<B<1.$$ --Lambiam 21:25, 19 June 2023 (UTC)
 * First, thank you for your response. Second, please notice I've inserted some words [in brackets] into your first sentence (after I addded some new paragraphs to my previous response), I hope it's ok (if it's not feel free to delete what I inserted into your first sentence). Third, you claim: "the equation $$f'(x)=g(x)$$ can also not have any positive solutions". But how can your claim fill in the blanks, in my paragraph #5? Some words should be inserted there, so what are they, according to your claim? 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 22:39, 19 June 2023 (UTC)
 * "it allows a (positive) solution of the equation $$f'(x)=f(x)/x$$". This is equivalent to: "the tangent to the graph of $$y=f(x),$$ for some $$x>0,$$ passes through the origin". --Lambiam 22:51, 19 June 2023 (UTC)
 * Do you mean, that every continuous function $$f,$$ having a (positive?) solution for the equation $$f'(x)=f(x)/x,$$ is an injection which satisfies that the function $$f(x)/x$$ is not an injection? Or you mean that every continuous injection $$f,$$ having a (positive?) solution for the equation $$f'(x)=f(x)/x,$$ satisfies that the function $$f(x)/x$$ is not an injection? Or you mean something different? 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 01:39, 20 June 2023 (UTC)
 * A solution of the equation, if it exists, is a value of $$x$$ satisfying the equation. It is positive if $$x>0.$$ The phrase was meant to fit in the slot of your #5, as requested. The assumption is that $$f$$ is a continuous (and even differentiable) injection defined on $$\R^+.$$ So,
 * $$f$$ cannot be any continuous (and even differentiable) injection defined on $$\R^+$$ having the following general (non-obvious) property: it allows a (positive) solution of the equation $$f'(x)=f(x)/x,$$ because even though $$f$$ is a continuous injection as required, $$g$$ is not.
 * --Lambiam 07:57, 20 June 2023 (UTC)
 * Thank you again, ever so much, for filling in the slot of my #5. Just to be sure, which one of the following sentences is necessarily true? Both?
 * Strong version: Every differentiable function $$f,$$ defined on $$\R^+$$, and allowing a positive solution of the equation $$f'(x)=f(x)/x,$$ is an injection which satisfies that the function $$f(x)/x$$ is not an injection.
 * Weak version: Every differentiable injection $$f,$$ defined on $$\R^+$$, and allowing a positive solution of the equation $$f'(x)=f(x)/x,$$ satisfies that the function $$f(x)/x$$ is not an injection.
 * Additionaly, what about the opposite direction of your claim: For every differentiable injection $$f, $$ defined on $$\R^+$$, and satisfying that $$f(x)/x$$ is not injective, is it true to claim that the equation $$f'(x)=f(x)/x$$ has a (positive) solution?
 * 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 12:24, 20 June 2023 (UTC)
 * The strong version is false. Counterexample: take $$f(x)=x^2-4x+9.$$ Then $$f'(3)=f(3)/3=2,$$ yet $$f(1)=f(3)=6.$$ I think both the weak version and its converse are valid: for any differentiable injection $$f$$ defined on $$\R^+,$$ function $$x\mapsto f(x)/x$$ is not injective if and only if $$f'(x)=f(x)/x$$ has a solution in $$\R^+.$$ --Lambiam 22:24, 20 June 2023 (UTC)
 * Thanx ever so much. 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 08:51, 21 June 2023 (UTC)
 * I'm sorry, but I have to retract part of this statement. Let $$f(x)=x^2+x\sin x.$$ Both $$f$$ and $$x\mapsto f(x)/x$$ are injective on $$\R^+,$$ but $$f'(x)=f(x)/x$$ for $$x=2k\pi,k\in \Z^+.$$ It appears that only the converse of the weak version is left standing. --Lambiam 15:18, 21 June 2023 (UTC)
 * Now you deserve a double thanx, because of your honesty. Honesty is the best policy. 2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 21:02, 21 June 2023 (UTC)